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1 1 1 1 --------- + --------- + --------- + ...+ ----------------- a-b a^2-b^2 a^4-b^4 a^(2^n)-b^(2^n) 想請問此級數和的前n項的表示法要怎麼求 懇請指點 感恩!!! -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 59.104.37.160
yueayase :不就sigma(1~n)(1/(a^(2^n)-b^(2^n)))? 12/04 00:22
ideaptt :1/(a-b)+1/(a^2-b^2)=[(a^2+b^2)+a+b+1]/(a^4-b^4) 12/05 01:32
ideaptt :correction 12/05 01:33
ideaptt :1/(a-b)+1/(a^2-b^2)=[(a^2+b^2)+a+b+1]/(a^2-b^2) 12/05 01:34
ideaptt :above+1/(a^4-b^4)=... 12/05 01:36
ideaptt :...=[(a^4+b^4)+a^2+b^2+a+b+1]/(a^4-b^4) 12/05 01:37
ideaptt :see the rule? move the dominator's part to the 12/05 01:39
ideaptt :numerator and the numerator could be separated 12/05 01:39
ideaptt :into the geometry seq. of a,b respectively. so 12/05 01:40
ideaptt :the numerator could expressed as a formula of n 12/05 01:41
ideaptt :and the dominator is simple. 12/05 01:41
sneak : into the ge http://yofuk.com 01/02 15:37
muxiv : 1/(a-b)+1/( https://moxox.com 07/07 11:41